Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $t = \dfrac{q^2 - 64}{-9q - 54} \times \dfrac{q + 6}{8q + 64} $
Solution: First factor the quadratic. $t = \dfrac{(q + 8)(q - 8)}{-9q - 54} \times \dfrac{q + 6}{8q + 64} $ Then factor out any other terms. $t = \dfrac{(q + 8)(q - 8)}{-9(q + 6)} \times \dfrac{q + 6}{8(q + 8)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ (q + 8)(q - 8) \times (q + 6) } { -9(q + 6) \times 8(q + 8) } $ $t = \dfrac{ (q + 8)(q - 8)(q + 6)}{ -72(q + 6)(q + 8)} $ Notice that $(q + 6)$ and $(q + 8)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ \cancel{(q + 8)}(q - 8)(q + 6)}{ -72(q + 6)\cancel{(q + 8)}} $ We are dividing by $q + 8$ , so $q + 8 \neq 0$ Therefore, $q \neq -8$ $t = \dfrac{ \cancel{(q + 8)}(q - 8)\cancel{(q + 6)}}{ -72\cancel{(q + 6)}\cancel{(q + 8)}} $ We are dividing by $q + 6$ , so $q + 6 \neq 0$ Therefore, $q \neq -6$ $t = \dfrac{q - 8}{-72} $ $t = \dfrac{-(q - 8)}{72} ; \space q \neq -8 ; \space q \neq -6 $